Re: My paper, and the cheaters

From: James Harris (jstevh_at_msn.com)
Date: 09/17/04 

Date: 17 Sep 2004 15:43:52 -0700

norabaron@hotmail.com (Nora Baron) wrote in message news:<36024859.0409170732.566a5e55@posting.google.com>...
> jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0409161404.365eaa1e@posting.google.com>...
> > I've made some pointed criticisms of math society but if you actually
> > pay attention you'll realize that the evidence against math society is
> > nastier than I go into detail about usually.
> >
> > Supposedly a correct math result is what's important, but despite my
> > having a quite correct paper, which was to be published, sci.math'ers
> > managed to get it yanked IN ONE DAY with some hostile emails.
> >
> > The paper has been by a lot of editors and mathematicians and to date,
> > no major error has been found, which I say because there was one minor
> > error that actually got pointed out by a sci.math poster.
> >
>
> Please do not ever again claim to be an honest person.
> Your "proof" contradicts known mathematics. At least 4
> counterproofs have been given. Dale Hall's counterproof
> can be checked with simple arithmetic. The exact spot
> in your "proof" where you make your major error has been pointed
> out repeatedly. You know all this and yet you continue to
> deny it.

That's not true.

In fact, I don't disagree with much of what you say.

And I'd like you to admit that in a post, so that I can move on to
what I actually *DO* say.

First though, I need you to read what I have below and acnowledge that
you see the agreement on a key point.

>
> Below is one of the simplest counterproofs. Feel free
> to point out errors (assuming you are still interested in
> math as opposed to social studies). This is, incidentally,
> for your future reference, what real proof is supposed to
> look like.
>
> ================================================================
>
> James Harris has written a paper called "Advanced Polynomial
> Factorization" in which he claims the following: if the
> polynomial
>
> 65*x^3 - 12*x + 1
>
> is factored in the form
>
> (a1*x + 1)*(a2*x + 1)*(a3*x + 1),
>
> where a1, a2 and a3 are algebraic integers, then exactly
> two of a1, a2, and a3 are divisible in the algebraic integers
> by sqrt(5), and the third one is coprime to 5.
>
> It should be noted that -a1, -a2, and -a3 are roots of
>
> x^3 + 12*x^2 - 65.
>
> That Harris's claim is false can be seen from the following:
>
> ==========================================================================
>
> Assume m(x) is an irreducible monic polynomial with integer coefficients
> and algebraic integers a and b are two roots of m(x).
>
> Theorem. If p is a nonzero rational integer and a is divisible by
> sqrt(p) in the algebraic integers, then b is also divisible
> by sqrt(p).

That last should say, in the algebraic integers.

Theorem accepted. Proof not needed here so I deleted it out to focus
on what's important. It is in fact true that if 'a' is divisible by
sqrt(p) in the algebraic integers then b is also divisible by sqrt(p)
***in the ring of algebraic integers***.

Concede agreement "Nora Baron" and then I'll explain the rest.

>
> This polynomial is monic and irreducible with integer coefficients.
> By the theorem above, if one of the roots is divisible by sqrt(5),
> then they all are. This is not possible because the product of the
> roots is 65 = 5 * 13. Therefore Harris's claim that ANY of the
> roots are divisible by sqrt(5) is false.

The poster has shifted from the true claim--that the result applies to
algebraic integers--to a far vaguer and more inclusive claim,
basically that it applies in general.

Repeatedly, posters keep making claims true in the ring of algebraic
integers, I accept those claims in that ring, and they then act like I
don't!!!

I repeat, the result "Nora Baron" gave is correct *in the ring of
algebraic integers* and I now challenge that poster and others in that
camp to finally and for once concede that I am not arguing against
that point.

Then I'd like, with the permission of "Nora Baron" to explain exactly
what my proof actually says.

But first I want the poster "Nora Baron" to accept that I am not
challenging on this specific point.

That's the first step.

It's up to that poster to respond.

James Harris

Relevant Pages

  • Re: JSH: Resolution now possible
    ... Very smart "Nora Baron". ... and constant multiples of polynomials are not functions, ... What I know is that constant terms are not varying functions. ... That is, given two algebraic integers a and b, ...
    (sci.math)
  • Re: My paper, and the cheaters
    ... the ring of algebraic integers" is not enough, ... >> It's up to that poster to respond. ... James Harris ...
    (sci.math)
  • Re: JSH:Understanding constant terms
    ... > not giving the ring of algebraic integers a special position. ... > Notice that the poster didn't say, in the ring of algebraic integers. ... > a constant term, which has to be made explicit in a bit. ...
    (sci.math)
  • Re: JSH:Understanding constant terms
    ... > not giving the ring of algebraic integers a special position. ... > Notice that the poster didn't say, in the ring of algebraic integers. ... > a constant term, which has to be made explicit in a bit. ...
    (sci.math)
  • Re: JSH:Understanding constant terms
    ... > not giving the ring of algebraic integers a special position. ... > Notice that the poster didn't say, in the ring of algebraic integers. ... > a constant term, which has to be made explicit in a bit. ...
    (sci.math)
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